Re: Need proof of simple result

On 22 Aug., 05:25, junoexpress <MTBrenne...@xxxxxxxxx> wrote:
Hi,

I am looking for a proof of the following statement:

Let x be a point in R^n.
Let S be a closed bounded convex set in R^n whose surface is smooth.
Then the point in S, x*, closest to x is the point s.t. x0-x is along
the surface normal of S at x*.

Does this result have a name and is there any texts were this is
proved? It seems obvious, but I haven't been able to find the proof.

TIA,

Matt

The statement is only true if x is not in S.

Since S is compact and distance is continuous, a closest
point x* exists.
It is unique because if y were another closest point, the point
(x* + y)/2 would be even closer and in S by convexity.

Because(!) x is not in S, we can consider the line xx*.
No point on xx* between x and x* belongs to S (by minimality
of dist(x,x*)).
Therefore x* cannot be an interior point and is on
the boundary dS.

Assume <y-x*,x-x*> >0 for some y in S.
Then some point on the line segment between x* and y is
closer to x than x* -- contradiction.
Therefore, <y-x*,x-x*> <= 0 for all y in S.
Therefore {y : <y-x*,x-x*> = 0} is a hyperplane that
is tangent to S at x*.
Since dS is smooth,there is only one such tangent
hyperplane, which is orthogonal to the normal at x*.
Thus x-x* is a (positive) multiple of the normal at x*.

hagman

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